>>475
∫[0,1]x^2dx
=∫[0,arctan(1)](tan(u))^2 * {du/(cos(u))^2}
=[(1/3)(tan(u))^3][0,arctan(1)]
=(1/3)tan(arctan(1))
=1/3

u:0→π/4+nπ でも変わらんよな。